
Re: Sum of primes divided by the next prime
Posted:
Aug 2, 2006 4:23 PM


You are right! AFAICS
Tapio
"Tim Peters" <tim.one@comcast.net> wrote in message news:beednTIHXN9BnkzZnZ2dnUVZ_vSdnZ2d@comcast.com... > [Tapio] >> Sum the primes up to p_n. Divide the sum by the next prime p_(n+1): >> >> Sum (2+3+...+p_n)/(p_(n+1)) >> >> I found with the aid of excel two examples: >> >> (2+3) = 0 mod 5, i.e. the next prime 5 divides the sum of all the >> preceding primes. > > While degenerate, it would also be reasonable to say that 2 divides 0. > >> (2+3+....+61) = 0 mod 67 > > Not when I do it ;) > > sum(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61) = > 501 = > 7*67 + 32 > > However, I do get that 71 divides 568. > >> Below the prime 10009 there is no other cases, except those mentioned >> above. >> Are there any other p_n > 10009 ? > > Through the 50,847,534 primes less than a billion (10^9), the only cases I > found were: > > 2 divides 0 > 5 divides 5 > 71 divides 568 > 369119 divides 5536785000 > 415074643 divides 4456254649636576 > > Best guess is that mod(sum(p_i, i = 1 to n), p_(n+1)) is "essentially > random", so that the chance of seeing 0 is 1/p_(n+1). Then since > sum(1/p_i) is divergent (albeit very slowly), best guess is that the > number of primes for which the mod is 0 is infinite (although their > density will decrease rapidly). Suggestive: > > p_n sum(1/p for prime p in [2, p_n]) >   > 5 1.03 > 71 1.74 > 369119 2.81 > 415074643 3.25 > >

